The Rocket Equation The Tsiolovsky Rocket Equation describes the velocity that results from pushing matter (exploding rocket fuel) in the opposite direction to the direction you want to travel. This assignment requires you to do basic calculation using the Tsiolovsky Rocket Equation : v[t] = eV Log M M – bR t  – g t The parameters used are : ◼ eV exhaust velocity (m/s) ◼ pL payload (kg) ◼ fL fuel load (kg) ◼ M is the mass of the rocket (pL+fL, kg) ◼ bR the burn rate of fuel (kg/s) ◼ g the force due to gravity ms2 The variables calculated are : h(t) the height of the rocket at time t (m) v(t) the velocity of the rocket at time t (m/s) m(t) the mass of the rocket at time t (kg) Questions Question 1 (1 mark) Write an expression corresponding to the Tsiolovsky rocket equation and use integrate to find a function to describe the height of the rocket during fuel burn. Question 2 (2 marks) The fuel burns at a constant rate. Find the time (t0), velocity (vmax), and height (h0) of the rocket when the fuel runs out (calculate the time when the fuel runs out, and substitute this into the height Printed by Wolfram Mathematica Student Edition and velocity equations). Question 3 (2 marks) The second phase is when the only accelaration acting on the rocket is from gravity. This phase starts from the height and velocity of the previous question, and the velocity is given by the projectile motion equation, v(t) = vmax – g (t – t0). Use Solve to find the time when this equation equals 0. This will be the highest point the rocket reaches before returning to earth. Question 4 (1 marks) Integerate the projectile motion equation and add h0 to find the maximum height the rocket reaches. Question 5 (1 marks) Use Solve over the projectile motion equation to find the time when the height is 0. 2 assignment4.nb Printed by Wolfram Mathematica Student Edition

The Rocket Equation The Tsiolovsky Rocket Equation describes the velocity that results from pushing matter (exploding rocket fuel) in the opposite direction to the direction you want to travel. This assignment requires you to do basic calculation using the Tsiolovsky Rocket Equation : v[t] = eV Log M M – bR t  – g t The parameters used are : ◼ eV exhaust velocity (m/s) ◼ pL payload (kg) ◼ fL fuel load (kg) ◼ M is the mass of the rocket (pL+fL, kg) ◼ bR the burn rate of fuel (kg/s) ◼ g the force due to gravity ms2 The variables calculated are : h(t) the height of the rocket at time t (m) v(t) the velocity of the rocket at time t (m/s) m(t) the mass of the rocket at time t (kg) Questions Question 1 (1 mark) Write an expression corresponding to the Tsiolovsky rocket equation and use integrate to find a function to describe the height of the rocket during fuel burn. Question 2 (2 marks) The fuel burns at a constant rate. Find the time (t0), velocity (vmax), and height (h0) of the rocket when the fuel runs out (calculate the time when the fuel runs out, and substitute this into the height Printed by Wolfram Mathematica Student Edition and velocity equations). Question 3 (2 marks) The second phase is when the only accelaration acting on the rocket is from gravity. This phase starts from the height and velocity of the previous question, and the velocity is given by the projectile motion equation, v(t) = vmax – g (t – t0). Use Solve to find the time when this equation equals 0. This will be the highest point the rocket reaches before returning to earth. Question 4 (1 marks) Integerate the projectile motion equation and add h0 to find the maximum height the rocket reaches. Question 5 (1 marks) Use Solve over the projectile motion equation to find the time when the height is 0. 2 assignment4.nb Printed by Wolfram Mathematica Student Edition

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COMP 4440/5440 – Dr. Erdemir Mobile Robotics Project (DUE 12/02/2015) HONOR CODE I pledge my honor that I have neither given nor received aid on this work. Do not sign until after you have completed your assignment. Name: Signature: 1. (Prerequisite) Given the asset package (it is in mytsu under assignments folder) download it, open a new project (don’t double click on the file, it won’t open), go to Assets/Import Package/Custom Package, select the asset package given to you (project.unitypackage). After the import is completed, you will see main scene in the assets folder of your project. Double click on it, and choose “Don’t Save” option if it asks for save. 2. Print this page and attach it to your code and your snapshot of your final scene (5 points) 3. After the class starts, instructor will come next to you and you are supposed to show him your code running (10 points) 4. When you run the code you will see, your robot (red cube that we used in the class) is trying to reach the targets but it can’t due to an obstacle between your robot and the targets. Write a code that makes this robot to avoid from the obstacles and reach the targets. Your code will read the collision and intelligently avoid from other moving robots and fixed obstacles and get the three targets. (50 points) 5. Maximum time is 2 minutes, your robot shall get the targets in 2 minutes (10 points) 6. Your robot shall escape from the blue robot and not collide them. (10 Points) 7. Anything extra (up to 20 points) ? Moving objects, new sensor, Artificial Intelligence or other techniques. 8. YOU CAN NOT USE TRANSLATE FUNCTION. USE ONLY AddRelativeForce FUNCTION IN THE FORWARD DIRECTION AS ALL THE MOBILE ROBOTS WORK.

COMP 4440/5440 – Dr. Erdemir Mobile Robotics Project (DUE 12/02/2015) HONOR CODE I pledge my honor that I have neither given nor received aid on this work. Do not sign until after you have completed your assignment. Name: Signature: 1. (Prerequisite) Given the asset package (it is in mytsu under assignments folder) download it, open a new project (don’t double click on the file, it won’t open), go to Assets/Import Package/Custom Package, select the asset package given to you (project.unitypackage). After the import is completed, you will see main scene in the assets folder of your project. Double click on it, and choose “Don’t Save” option if it asks for save. 2. Print this page and attach it to your code and your snapshot of your final scene (5 points) 3. After the class starts, instructor will come next to you and you are supposed to show him your code running (10 points) 4. When you run the code you will see, your robot (red cube that we used in the class) is trying to reach the targets but it can’t due to an obstacle between your robot and the targets. Write a code that makes this robot to avoid from the obstacles and reach the targets. Your code will read the collision and intelligently avoid from other moving robots and fixed obstacles and get the three targets. (50 points) 5. Maximum time is 2 minutes, your robot shall get the targets in 2 minutes (10 points) 6. Your robot shall escape from the blue robot and not collide them. (10 Points) 7. Anything extra (up to 20 points) ? Moving objects, new sensor, Artificial Intelligence or other techniques. 8. YOU CAN NOT USE TRANSLATE FUNCTION. USE ONLY AddRelativeForce FUNCTION IN THE FORWARD DIRECTION AS ALL THE MOBILE ROBOTS WORK.

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Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

Extra Credit Due: 11:59pm on Thursday, May 15, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy A Man Running to Catch a Bus A man is running at speed (much less than the speed of light) to catch a bus already at a stop. At , when he is a distance from the door to the bus, the bus starts moving with the positive acceleration . Use a coordinate system with at the door of the stopped bus. Part A What is , the position of the man as a function of time? Answer symbolically in terms of the variables , , and . Hint 1. Which equation should you use for the man’s speed? Because the man’s speed is constant, you may use . ANSWER: c t = 0 b a x = 0 xman(t) b c t x(t) = x(0) + vt xman(t) = −b + ct Correct Part B What is , the position of the bus as a function of time? Answer symbolically in terms of and . Hint 1. Which equation should you use for the bus’s acceleration? Because the bus has constant acceleration, you may use . Recall that . ANSWER: Correct Part C What condition is necessary for the man to catch the bus? Assume he catches it at time . Hint 1. How to approach this problem If the man is to catch the bus, then at some moment in time , the man must arrive at the position of the door of the bus. How would you express this condition mathematically? ANSWER: xbus(t) a t x(t) = x(0) + v(0)t + (1/2)at2 vbus(0) = 0 xbus = 1 a 2 t2 tcatch tcatch Typesetting math: 15% Correct Part D Inserting the formulas you found for and into the condition , you obtain the following: , or . Intuitively, the man will not catch the bus unless he is running fast enough. In mathematical terms, there is a constraint on the man’s speed so that the equation above gives a solution for that is a real positive number. Find , the minimum value of for which the man will catch the bus. Express the minimum value for the man’s speed in terms of and . Hint 1. Consider the discriminant Use the quadratic equation to solve: . What is the discriminant (the part under the radical) of the solution for ? xman(tcatch) > xbus(tcatch) xman(tcatch) = xbus(tcatch) xman(tcatch) < xbus(tcatch) c = a  tcatch xman(t) xbus(t) xman(tcatch) = xbus(tcatch) −b+ct = a catch 1 2 t2 catch 1 a −c +b = 0 2 t2 catch tcatch c tcatch cmin c a b 1 a − c + b = 0 2 t2 catch tcatch tcatch Typesetting math: 15% Hint 1. The quadratic formula Recall: If then ANSWER: Hint 2. What is the constraint? To get a real value for , the discriminant must be greater then or equal to zero. This condition yields a constraint that exceed . ANSWER: Correct Part E Assume that the man misses getting aboard when he first meets up with the bus. Does he get a second chance if he continues to run at the constant speed ? Hint 1. What is the general quadratic equation? The general quadratic equation is , where , \texttip{B}{B}, and \texttip{C}{C} are constants. Depending on the value of the discriminant, \Delta = c^2-2ab, the equation may have Ax2 + Bx + C = 0 x = −B±B2−4AC 2A  = cc − 2ab tcatch c cmin cmin = (2ab) −−−−  c > cmin Ax2 + Bx + C = 0 A Typesetting math: 15% two real valued solutions 1. if \Delta > 0, 2. one real valued solution if \Delta = 0, or 3. two complex valued solutions if \Delta < 0. In this case, every real valued solution corresponds to a time at which the man is at the same position as the door of the bus. ANSWER: Correct Adding and Subtracting Vectors Conceptual Question Six vectors (A to F) have the magnitudes and directions indicated in the figure. Part A No; there is no chance he is going to get aboard. Yes; he will get a second chance Typesetting math: 15% Which two vectors, when added, will have the largest (positive) x component? Hint 1. Largest x component The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive x components are larger than negative x components. ANSWER: Correct Part B Which two vectors, when added, will have the largest (positive) y component? Hint 1. Largest y component The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive y components are larger than negative y components. ANSWER: C and E E and F A and F C and D B and D Typesetting math: 15% Correct Part C Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude? Hint 1. Subtracting vectors To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector. ANSWER: Correct Tactics Box 3.1 Determining the Components of a Vector Learning Goal: C and D A and F E and F A and B E and D A and F A and E D and B C and D E and F Typesetting math: 15% To practice Tactics Box 3.1 Determining the Components of a Vector. When a vector \texttip{\vec{A}}{A_vec} is decomposed into component vectors \texttip{\vec{A}_{\mit x}}{A_vec_x} and \texttip{\vec{A}_{\mit y}}{A_vec_y} parallel to the coordinate axes, we can describe each component vector with a single number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector \texttip{\vec{A}}{A_vec}, denoted \texttip{A_{\mit x}}{A_x} and \texttip{A_{\mit y}}{A_y}. TACTICS BOX 3.1 Determining the components of a vector The absolute value |A_x| of the x component \texttip{A_{\mit x}}{A_x} is the magnitude of the component vector \texttip{\vec{A}_{\1. mit x}}{A_vec_x}. The sign of \texttip{A_{\mit x}}{A_x} is positive if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the positive x direction; it is negative if \texttip{\vec{A}_{\mit x}}{A_vec_x} points in the negative x direction. 2. 3. The y component \texttip{A_{\mit y}}{A_y} is determined similarly. Part A What is the magnitude of the component vector \texttip{\vec{A}_{\mit x}}{A_vec_x} shown in the figure? Express your answer in meters to one significant figure. ANSWER: Correct |A_x| = 5 \rm m Typesetting math: 15% Part B What is the sign of the y component \texttip{A_{\mit y}}{A_y} of vector \texttip{\vec{A}}{A_vec} shown in the figure? ANSWER: Correct Part C Now, combine the information given in the tactics box above to find the x and y components, \texttip{B_{\mit x}}{B_x} and \texttip{B_{\mit y}}{B_y}, of vector \texttip{\vec{B}}{B_vec} shown in the figure. Express your answers, separated by a comma, in meters to one significant figure. positive negative Typesetting math: 15% ANSWER: Correct Conceptual Problem about Projectile Motion Learning Goal: To understand projectile motion by considering horizontal constant velocity motion and vertical constant acceleration motion independently. Projectile motion refers to the motion of unpowered objects (called projectiles) such as balls or stones moving near the surface of the earth under the influence of the earth's gravity alone. In this analysis we assume that air resistance can be neglected. An object undergoing projectile motion near the surface of the earth obeys the following rules: An object undergoing projectile motion travels horizontally at a constant rate. That is, the x component of its velocity, \texttip{v_{\mit x}}{1. v_x}, is constant. An object undergoing projectile motion moves vertically with a constant downward acceleration whose magnitude, denoted by \texttip{g}{g}, is equal to 9.80 \rm{m/s^2} near the surface of the earth. Hence, the y component of its velocity, \texttip{v_{\mit y}}{v_y}, changes continuously. 2. An object undergoing projectile motion will undergo the horizontal and vertical motions described above from the instant it is launched until the instant it strikes the ground again. Even though the horizontal and vertical motions can be treated independently, they are related by the fact that they occur for exactly the same amount of time, namely the time \texttip{t}{t} the projectile is in the air. 3. The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0 = 0\;\rm{s} corresponds to the moment just after the ball is launched from position x_0 = 0\;\rm{m} and y_0 = 0\;\rm{m}. Its launch velocity, also called the initial velocity, is \texttip{\vec{v}_{\rm 0}}{v_vec_0}. Two other points along the trajectory are indicated in the figure. One is the moment the ball reaches the peak of its trajectory, at time \texttip{t_{\rm 1}}{t_1} with velocity \texttip{\vec{v}_{\rm 1}}{v_1_vec}. Its position at this moment is denoted by (x_1, y_1) or (x_1, y_{\max}) since it is at its maximum \texttip{B_{\mit x}}{B_x}, \texttip{B_{\mit y}}{B_y} = -2,-5 \rm m, \rm m Typesetting math: 15% The other point, at time \texttip{t_{\rm 2}}{t_2} with velocity \texttip{\vec{v}_{\rm 2}}{v_2_vec}, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is (x_2, y_2), also known as (x_{\max}, y_2) since it is at its maximum horizontal range. Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which is was launched, as is the case here. Hence y_2 = y_0 = 0\;\rm{m}. Part A How do the speeds \texttip{v_{\rm 0}}{v_0}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2} (at times \texttip{t_{\rm 0}}{t_0}, \texttip{t_{\rm 1}}{t_1}, and \texttip{t_{\rm 2}}{t_2}) compare? ANSWER: Correct Here \texttip{v_{\rm 0}}{v_0} equals \texttip{v_{\rm 2}}{v_2} by symmetry and both exceed \texttip{v_{\rm 1}}{v_1}. This is because \texttip{v_{\rm 0}}{v_0} and \texttip{v_{\rm 2}}{v_2} include vertical speed as well as the constant horizontal speed. Consider a diagram of the ball at time \texttip{t_{\rm 0}}{t_0}. Recall that \texttip{t_{\rm 0}}{t_0} refers to the instant just after the ball has been launched, so it is still at ground level (x_0 = y_0= 0\;\rm{m}). However, it is already moving with initial velocity \texttip{\vec{v}_{\rm 0}}{v_0_vec}, whose magnitude is v_0 = 30.0\;{\rm m/s} and direction is \theta = 60.0\;{\rm degrees} counterclockwise from the positive x direction. \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 1}}{v_1} = \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 \texttip{v_{\rm 0}}{v_0} = \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 1}}{v_1} > \texttip{v_{\rm 2}}{v_2} > 0 \texttip{v_{\rm 0}}{v_0} > \texttip{v_{\rm 2}}{v_2} > \texttip{v_{\rm 1}}{v_1} = 0 Typesetting math: 15% Part B What are the values of the intial velocity vector components \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{0,x}}{a_0, x} and \texttip{a_{0,y}}{a_0, y} (both in \rm{m/s^2})? Here the subscript 0 means “at time \texttip{t_{\rm 0}}{t_0}.” Hint 1. Determining components of a vector that is aligned with an axis If a vector points along a single axis direction, such as in the positive x direction, its x component will be its full magnitude, whereas its y component will be zero since the vector is perpendicular to the y direction. If the vector points in the negative x direction, its x component will be the negative of its full magnitude. Hint 2. Calculating the components of the initial velocity Notice that the vector \texttip{\vec{v}_{\rm 0}}{v_0_vec} points up and to the right. Since “up” is the positive y axis direction and “to the right” is the positive x axis direction, \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} will both be positive. As shown in the figure, \texttip{v_{0,x}}{v_0, x}, \texttip{v_{0,y}}{v_0, y}, and \texttip{v_{\rm 0}}{v_0} are three sides of a right triangle, one angle of which is \texttip{\theta }{theta}. Thus \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y} can be found using the definition of the sine and cosine functions given below. Recall that v_0 = 30.0\;\rm{m/s} and \theta = 60.0\;\rm{degrees} and note that \large{\sin(\theta) = \frac{\rm{length\;of\;opposite\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, y}}{v_0}}, \large{\cos(\theta) = \frac{\rm{length\;of\;adjacent\;side}}{\rm{length\;of\;hypotenuse}}} \large{= \frac{v_{0, x}}{v_0}.} What are the values of \texttip{v_{0,x}}{v_0, x} and \texttip{v_{0,y}}{v_0, y}? Enter your answers numerically in meters per second separated by a comma. ANSWER: ANSWER: 15.0,26.0 \rm{m/s} Typesetting math: 15% Correct Also notice that at time \texttip{t_{\rm 2}}{t_2}, just before the ball lands, its velocity components are v_{2, x} = 15\;\rm{m/s} (the same as always) and v_{2, y} = – 26.0\;\rm{m/s} (the same size but opposite sign from \texttip{v_{0,y}}{v_0, y} by symmetry). The acceleration at time \texttip{t_{\rm 2}}{t_2} will have components (0, -9.80 \rm{m/s^2}), exactly the same as at \texttip{t_{\rm 0}}{t_0}, as required by Rule 2. The peak of the trajectory occurs at time \texttip{t_{\rm 1}}{t_1}. This is the point where the ball reaches its maximum height \texttip{y_{\rm max}}{y_max}. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate. Part C What are the values of the velocity vector components \texttip{v_{1,x}}{v_1, x} and \texttip{v_{1,y}}{v_1, y} (both in \rm{m/s}) as well as the acceleration vector components \texttip{a_{1,x}}{a_1, x} and \texttip{a_{1,y}}{a_1, y} (both in \rm{m/s^2})? Here the subscript 1 means that these are all at time \texttip{t_{\rm 1}}{t_1}. ANSWER: 30.0, 0, 0, 0 0, 30.0, 0, 0 15.0, 26.0, 0, 0 30.0, 0, 0, -9.80 0, 30.0, 0, -9.80 15.0, 26.0, 0, -9.80 15.0, 26.0, 0, +9.80 Typesetting math: 15% Correct At the peak of its trajectory the ball continues traveling horizontally at a constant rate. However, at this moment it stops moving up and is about to move back down. This constitutes a downward-directed change in velocity, so the ball is accelerating downward even at the peak. The flight time refers to the total amount of time the ball is in the air, from just after it is launched (\texttip{t_{\rm 0}}{t_0}) until just before it lands (\texttip{t_{\rm 2}}{t_2}). Hence the flight time can be calculated as t_2 – t_0, or just \texttip{t_{\rm 2}}{t_2} in this particular situation since t_0 = 0. Because the ball lands at the same height from which it was launched, by symmetry it spends half its flight time traveling up to the peak and the other half traveling back down. The flight time is determined by the initial vertical component of the velocity and by the acceleration. The flight time does not depend on whether the object is moving horizontally while it is in the air. Part D If a second ball were dropped from rest from height \texttip{y_{\rm max}}{y_max}, how long would it take to reach the ground? Ignore air resistance. Check all that apply. Hint 1. Kicking a ball of cliff; a related problem Consider two balls, one of which is dropped from rest off the edge of a cliff at the same moment that the other is kicked horizontally off the edge of the cliff. Which ball reaches the level ground at the base of the cliff first? Ignore air resistance. Hint 1. Comparing position, velocity, and acceleration of the two balls Both balls start at the same height and have the same initial y velocity (v_{0,y} = 0) as well as the same acceleration (\vec a = g downward). They differ only in their x velocity (one is 0, 0, 0, 0 0, 0, 0, -9.80 15.0, 0, 0, 0 15.0, 0, 0, -9.80 0, 26.0, 0, 0 0, 26.0, 0, -9.80 15.0, 26.0, 0, 0 15.0, 26.0, 0, -9.80 Typesetting math: 15% zero, the other nonzero). This difference will affect their x motion but not their y motion. ANSWER: ANSWER: Correct In projectile motion over level ground, it takes an object just as long to rise from the ground to the peak as it takes for it to fall from the peak back to the ground. The range \texttip{R}{R} of the ball refers to how far it moves horizontally, from just after it is launched until just before it lands. Range is defined as x_2 – x_0, or just \texttip{x_{\rm 2}}{x_2} in this particular situation since x_0 = 0. Range can be calculated as the product of the flight time \texttip{t_{\rm 2}}{t_2} and the x component of the velocity \texttip{v_{\mit x}}{v_x} (which is the same at all times, so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can be found from the launch speed \texttip{v_{\rm 0}}{v_0} and the launch angle \texttip{\theta }{theta} using trigonometric functions, as was done in Part B. The flight time is related to the initial y component of the velocity, which may also be found from \texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig functions. The following equations may be useful in solving projectile motion problems, but these equations apply only to a projectile launched over level ground from position (x_0 = y_0 = 0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and launch angle \texttip{\theta }{theta} measured from the horizontal. As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the flight time and \texttip{R}{R} refers to the range of the projectile. flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0 \sin(\theta)}{g}} range: \large{R = v_x t_2 = \frac{v_0^2 \sin(2\theta)}{g}} The ball that falls straight down strikes the ground first. The ball that was kicked so it moves horizontally as it falls strikes the ground first. Both balls strike the ground at the same time. \texttip{t_{\rm 0}}{t_0} t_1 – t_0 \texttip{t_{\rm 2}}{t_2} t_2 – t_1 \large{\frac{t_2 – t_0}{2}} Typesetting math: 15% In general, a high launch angle yields a long flight time but a small horizontal speed and hence little range. A low launch angle gives a larger horizontal speed, but less flight time in which to accumulate range. The launch angle that achieves the maximum range for projectile motion over level ground is 45 degrees. Part E Which of the following changes would increase the range of the ball shown in the original figure? Check all that apply. ANSWER: Correct A solid understanding of the concepts of projectile motion will take you far, including giving you additional insight into the solution of projectile motion problems numerically. Even when the object does not land at the same height from which is was launched, the rules given in the introduction will still be useful. Recall that air resistance is assumed to be negligible here, so this projectile motion analysis may not be the best choice for describing things like frisbees or feathers, whose motion is strongly influenced by air. The value of the gravitational free-fall acceleration \texttip{g}{g} is also assumed to be constant, which may not be appropriate for objects that move vertically through distances of hundreds of kilometers, like rockets or missiles. However, for problems that involve relatively dense projectiles moving close to the surface of the earth, these assumptions are reasonable. A World-Class Sprinter World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude 15 \;{\rm m}/{\rm s}^{2}. Part A How much horizontal force \texttip{F}{F} must a sprinter of mass 64{\rm kg} exert on the starting blocks to produce this acceleration? Express your answer in newtons using two significant figures. Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}. Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45 \rm{degrees}. Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less than 30 \rm{degrees}. Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward 90 \rm{degrees}. Typesetting math: 15% Hint 1. Newton’s 2nd law of motion According to Newton’s 2nd law of motion, if a net external force \texttip{F_{\rm net}}{F_net} acts on a body, the body accelerates, and the net force is equal to the mass \texttip{m}{m} of the body times the acceleration \texttip{a}{a} of the body: F_{\rm net} = ma. ANSWER: Co

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Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

Assignment 3 Due: 11:59pm on Friday, February 14, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 2.68 As a science project, you drop a watermelon off the top of the Empire State Building, 320 m above the sidewalk. It so happens that Superman flies by at the instant you release the watermelon. Superman is headed straight down with a speed of 36.0 . Part A How fast is the watermelon going when it passes Superman? Express your answer with the appropriate units. ANSWER: Correct Problem 2.63 A motorist is driving at when she sees that a traffic light ahead has just turned red. She knows that this light stays red for , and she wants to reach the light just as it turns green again. It takes her to step on the brakes and begin slowing. Part A What is her speed as she reaches the light at the instant it turns green? Express your answer with the appropriate units. ANSWER: m/s 72.0 ms 20 m/s 200 m 15 s 1.0 s 5.71 ms Correct Conceptual Question 4.1 Part A At this instant, is the particle in the figurespeeding up, slowing down, or traveling at constant speed? ANSWER: Correct Part B Is this particle curving to the right, curving to the left, or traveling straight? Speeding up Slowing down Traveling at constant speed ANSWER: Correct Conceptual Question 4.2 Part A At this instant, is the particle in the following figure speeding up, slowing down, or traveling at constant speed? ANSWER: Curving to the right Curving to the left Traveling straight Correct Part B Is this particle curving upward, curving downward, or traveling straight? ANSWER: Correct Problem 4.8 A particle’s trajectory is described by and , where is in s. Part A What is the particle’s speed at ? ANSWER: The particle is speeding up. The particle is slowing down. The particle is traveling at constant speed. The particle is curving upward. The particle is curving downward. The particle is traveling straight. x = ( 1 −2 ) m 2 t3 t2 y = ( 1 −2t) m 2 t2 t t = 0 s v = 2 m/s Correct Part B What is the particle’s speed at ? Express your answer using two significant figures. ANSWER: Correct Part C What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: Correct Part D What is the particle’s direction of motion, measured as an angle from the x-axis, at ? Express your answer using two significant figures. ANSWER: t = 5.0s v = 18 m/s t = 0 s  = -90  counterclockwise from the +x axis. t = 5.0s  = 9.7  counterclockwise from the +x axis. Correct Problem 4.9 A rocket-powered hockey puck moves on a horizontal frictionless table. The figure shows the graph of and the figure shows the graph of , the x- and y-components of the puck’s velocity, respectively. The puck starts at the origin. Part A In which direction is the puck moving at = 3 ? Give your answer as an angle from the x-axis. Express your answer using two significant figures. ANSWER: Correct Part B vx vy t s = 51   above the x-axis How far from the origin is the puck at 5 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.13 A rifle is aimed horizontally at a target 51.0 away. The bullet hits the target 1.50 below the aim point. You may want to review ( pages 91 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A What was the bullet’s flight time? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the bullet’s trajectory, including where it leaves the gun and where it hits the target. You can assume that the gun was held parallel to the ground. Label the distances given in the problem. Choose an x-y coordinate system, making sure to label the origin. It is conventional to have x in the horizontal direction and y in the vertical direction. What is the y coordinate when the bullet leaves the gun? What is the y coordinate when it hits the target? What is the initial velocity in the y direction? What is the acceleration in the y direction? What is the equation that describes the motion in the vertical y direction as a function of time? Can you use the equation for to determine the time of flight? Why was it not necessary to include the motion in the x direction? s s = 180 cm m cm y(t) y(t) ANSWER: Correct Part B What was the bullet’s speed as it left the barrel? Express your answer with the appropriate units. Hint 1. How to approach the problem In the coordinate system introduced in Part A, what are the x coordinates when the bullet leaves the gun and when it hits the target? Is there any acceleration in the x direction? What is the equation that describes the motion in the horizontal x direction as a function of time? Can you use the equation for to determine the initial velocity? ANSWER: Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really no different than the one-dimensional motions that you have already studied. One of the most often used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction on the xy plane and can have any magnitude. To make a problem more managable, it is common to break up such a quantity into its x component and its y component . 5.53×10−2 s x(t) x(t) 922 ms v vx vy Consider a particle with initial velocity that has magnitude 12.0 and is directed 60.0 above the negative x axis. Part A What is the x component of ? Express your answer in meters per second. ANSWER: Correct Part B What is the y component of ? Express your answer in meters per second. ANSWER: Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Look at this applet. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x-component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: v m/s degrees vx v vx = -6.00 m/s vy v vy = 10.4 m/s Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider this applet. Two balls are simultaneously dropped from a height of 5.0 . Part D How long does it take for the balls to reach the ground? Use 10 for the magnitude of the acceleration due to gravity. Express your answer in seconds to two significant figures. Hint 1. How to approach the problem The balls are released from rest at a height of 5.0 at time . Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground. ANSWER: Correct This situation, which you have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion where the horizontal component of the initial velocity is zero. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. g = 10 m/s2 y = y0 + v0 t + (1/2)at2 x = x0 + v0 t m tg m/s2 m t = 0 s tg = 1.0 s Part E Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Remember that the two balls are released, starting a horizontal distance of 3.0 apart. Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that, in a time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: Correct You can adjust the horizontal speeds in this applet. Notice that regardless of what horizontal speeds you give to the balls, they continue to move vertically in the same way (i.e., they are at the same y coordinate at the same time). Problem 4.12 A ball thrown horizontally at 27 travels a horizontal distance of 49 before hitting the ground. Part A From what height was the ball thrown? Express your answer using two significant figures with the appropriate units. ANSWER: vx m vx tg = 1.0 s m x0 = 0.0 m vx = 3.0 m/s m/s m h = 16 m Correct Enhanced EOC: Problem 4.20 The figure shows the angular-velocity-versus-time graph for a particle moving in a circle. You may want to review ( page ) . For help with math skills, you may want to review: The Definite Integral Part A How many revolutions does the object make during the first 3.5 ? Express your answer using two significant figures. You did not open hints for this part. ANSWER: s n = Incorrect; Try Again Problem 4.26 To withstand “g-forces” of up to 10 g’s, caused by suddenly pulling out of a steep dive, fighter jet pilots train on a “human centrifuge.” 10 g’s is an acceleration of . Part A If the length of the centrifuge arm is 10.0 , at what speed is the rider moving when she experiences 10 g’s? Express your answer with the appropriate units. ANSWER: Correct Problem 4.28 Your roommate is working on his bicycle and has the bike upside down. He spins the 60.0 -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. Part A What is the pebble’s speed? Express your answer with the appropriate units. ANSWER: Correct 98 m/s2 m 31.3 ms cm 5.65 ms Part B What is the pebble’s acceleration? Express your answer with the appropriate units. ANSWER: Correct Enhanced EOC: Problem 4.43 On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 13 at an angle 50 above the horizontal. You may want to review ( pages 90 – 95) . For help with math skills, you may want to review: Quadratic Equations Part A How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the golf ball, showing its starting and ending points. Choose a coordinate system, and label the origin. It is conventional to let x be the horizontal direction and y the vertical direction. What is the initial velocity in the x and y directions? What is the acceleration in the x and y directions on the moon and on the earth? What are the equations for and as a function of time, and , respectively? What is the y coordinate when the golf ball hits the ground? Can you use this information to determine the time of flight on the moon and on the earth? 107 m s2 m/s  x y x(t) y(t) Once you have the time of flight, how can you use the equation to determine the total distance traveled? Compare the distance traveled on the moon to the distance traveled on the earth . ANSWER: Correct Part B For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the equation describing as a function of time? What is the initial x component of the ball’s velocity? How are the initial x component of the ball’s velocity and the distance traveled related to the time of flight? What is the difference between the time of flight on the moon and on earth? ANSWER: Correct Problem 4.42 In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 at a 40.0 angle from the horizontal. The shot leaves her hand at a height of 1.8 above the ground. x(t) L = 85 m x(t) x t = 10 s m/s  m Part A How far does the shot travel? Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part B Repeat the calculation of part (a) for angles of 42.5 , 45.0 , and 47.5 . Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part C Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part D x = 16.36 m    x(42.5 ) = 16.39 m x(45.0 ) = 16.31 m Express your answer to four significant figures and include the appropriate units. ANSWER: Correct Part E At what angle of release does she throw the farthest? ANSWER: Correct Problem 4.44 A ball is thrown toward a cliff of height with a speed of 32 and an angle of 60 above horizontal. It lands on the edge of the cliff 3.2 later. Part A How high is the cliff? Express your answer to two significant figures and include the appropriate units. ANSWER: x(47.5 ) = 16.13 m 40.0 42.5 45.0 47.5 h m/s  s h = 39 m Answer Requested Part B What was the maximum height of the ball? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the ball’s impact speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 4.58 A typical laboratory centrifuge rotates at 3600 . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 from the axis of rotation? Express your answer with the appropriate units. hmax = 39 m v = 16 ms rpm cm ANSWER: Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 and stopped in a 1.7-ms-long encounter with a hard floor? Express your answer with the appropriate units. ANSWER: Correct Problem 4.62 Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The radius of the earth is , and the altitude of a geosynchronous orbit is ( 22000 miles). Part A What is the speed of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct a = 1.42×104 m s2 m a = 2610 m s2 6.37 × 106m 3.58 × 107m  v = 3070 ms Part B What is the magnitude of the acceleration of a satellite in a geosynchronous orbit? Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 89.5%. You received 103.82 out of a possible total of 116 points. a = 0.223 m s2

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Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

Assignment 2 Due: 11:59pm on Wednesday, February 12, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 2.6 Part A The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Is the object moving the slowest? Is the object moving the fastest? Is the object at rest? Drag the appropriate items to their respective bins. ANSWER: Correct Part B At which lettered point or points is the object moving to the negative direction? ANSWER: Correct Conceptual Question 2.7 The figure shows the position-versus-time graph for a moving object. At which lettered point or points: Part A Is the object moving the fastest? ANSWER: A B C D E Correct Part B Is the object speeding up? ANSWER: Correct Part C Is the object moving to the left and turning around? ANSWER: A B C D E F A B C D E F Correct Kinematic Vocabulary One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings, among them velocity, acceleration, position, speed, and displacement. The series of questions in this problem is designed to get you to try to think of these quantities like a physicist. Answer the questions in this problem using words from the following list: A. position B. direction C. displacement D. coordinates E. velocity F. acceleration G. distance H. magnitude I. vector J. scalar K. components Part A Velocity differs from speed in that velocity indicates a particle’s __________ of motion. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part B Unlike speed, velocity is a __________ quantity. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part C A vector has, by definition, both __________ and direction. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part D Once you have selected a coordinate system, you can express a two-dimensional vector using a pair of quantities known collectively as __________. Enter the letter from the list given in the problem introduction that best completes the sentence. ANSWER: Correct Part E Speed differs from velocity in the same way that __________ differs from displacement. Enter the letter from the list given in the problem introduction that best completes the sentence. Hint 1. Definition of displacement Displacement is the vector that indicates the difference of two positions (e.g., the final position from the initial position). Being a vector, it is independent of the coordinate system used to describe it (although its vector components depend on the coordinate system). ANSWER: Correct Part F Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate systems that are identical except that they have different origins. The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems. Type the letters from the list given in the problem introduction that best complete the sentence. Separate the letters with commas. There is more than one correct answer, but you should only enter one pair of comma-separated letters. For example, if the words “vector” and “scalar” fit best in the blanks, enter I,J. ANSWER: Correct The coordinates of a point will depend on the coordinate system that is chosen, but there are several other quantities that are independent of the choice of origin for a coordinate system: in particular, distance, displacement, direction, and velocity. In working physics problems, unless you are interested in the position of an object or event relative to a specific origin, you can usually choose the coordinate system origin to be wherever is most convenient or intuitive. Note that the vector indicating a displacement from A to B is usually represented as . Part G Identify the following physical quantities as scalars or vectors. ANSWER: rB A = rB − rA Correct Problem 2.4 The figure is the position-versus-time graph of a jogger. Part A What is the jogger’s velocity at = 10 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Part B What is the jogger’s velocity at = 25 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the jogger’s velocity at = 35 ? Express your answer to two significant figures and include the appropriate units. ANSWER: t s v = 1.3 ms t s v = 0 ms t s v = -5.0 ms Correct Analyzing Position versus Time Graphs: Conceptual Question Two cars travel on the parallel lanes of a two-lane road. The cars’ motions are represented by the position versus time graph shown in the figure. Answer the questions using the times from the graph indicated by letters. Part A At which of the times do the two cars pass each other? Hint 1. Two cars passing Two objects can pass each other only if they have the same position at the same time. ANSWER: Correct Part B Are the two cars traveling in the same direction when they pass each other? ANSWER: Correct Part C At which of the lettered times, if any, does car #1 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E None Cannot be determined yes no Correct Part D At which of the lettered times, if any, does car #2 momentarily stop? Hint 1. Determining velocity from a position versus time graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: A B C D E none cannot be determined A B C D E none cannot be determined Correct Part E At which of the lettered times are the cars moving with nearly identical velocity? Hint 1. Determining Velocity from a Position versus Time Graph The slope on a position versus time graph is the “rise” (change in position) over the “run” (change in time). In physics, the ratio of change in position over change in time is defined as the velocity. Thus, the slope on a position versus time graph is the velocity of the object being graphed. ANSWER: Correct Problem 2.6 A particle starts from 10 at = 0 and moves with the velocity graph shown in the figure. A B C D E None Cannot be determined m t0 Part A Does this particle have a turning point? ANSWER: Correct Part B If so, at what time? Express your answer using two significant figures and include the appropriate units. ANSWER: Correct Part C What is the object’s position at = 2, 3, 4 ? Yes No t = 1.0 s t s Express your answers using two significant figures separated by commas. ANSWER: Correct Overcoming a Head Start Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance beyond the starting line at . The starting line is at . Car A travels at a constant speed . Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed , which is greater than . Part A How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities. Hint 1. Consider the kinematics relation Write an expression for the displacement of Car A from the starting line at a time after Car B starts. (Note that we are taking this time to be .) Answer in terms of , , , and for time, and take at the starting line. Hint 1. What is the acceleration of Car A? The acceleration of Car A is zero, so the general formula has at least one term equal to zero. ANSWER: Hint 2. What is the relation between the positions of the two cars? x2 , x3 , x4 = 10,16,26 m DA t = 0 x = 0 vA vB vA t t = 0 vA vB DA t x = 0 x(t) = x0 + v0t + (1/2)at2 xA(t) = DA + vAt The positions of the two cars are equal at time . Hint 3. Consider Car B’s position as a function of time Write down an expression for the position of Car B at time after starting. Give your answer in terms of any variables needed (use for time). ANSWER: ANSWER: Correct Part B How far from Car B’s starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities. (You may use as well.) Hint 1. Which expression should you use? Just use your expression for the position of either car after time , and substitute in the correct value for (found in the previous part). ANSWER: Correct tcatch t t xB(t) = vBt tcatch = DA vB−vA tcatch t = 0 tcatch dpass = vBDA vB−vA Problem 2.11 The figure shows the velocity graph of a particle moving along the x-axis. Its initial position is at . At = 2 , what are the particle’s (a) position, (b) velocity, and (c) acceleration? Part A Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Express your answer to two significant figures and include the appropriate units. ANSWER: x0 = 2 m t0 = 0 t s x = 6.0 m vx = 4.0 ms Correct Part C Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 2.13 A jet plane is cruising at 300 when suddenly the pilot turns the engines up to full throttle. After traveling 3.9 , the jet is moving with a speed of 400 . Part A What is the jet’s acceleration, assuming it to be a constant acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 2.20 A rock is tossed straight up with a velocity of 22 When it returns, it falls into a hole deep. You may want to review ( pages 51 – 54) . ax = 2.0 m s2 m/s km m/s a = 9.0 m s2 m/s 10 m For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A What is the rock’s velocity as it hits the bottom of the hole? Express your answer with the appropriate units. Hint 1. How to approach the problem Start by drawing a picture of the path of the rock, including its launch point, initial direction, and end point in the hole. Choose a coordinate system, and indicate it on your picture. Where is ? What is the positive direction? What is the position of the launch point and the bottom of the hole? In this coordinate system, what is the sign of the initial velocity and the sign of the acceleration? Calling the launch time , what is the equation for as a function of time? What is the position at the bottom of the hole? This will lead to a quadratic equation for the time when the rock hits the bottom of the hole. The quadratic equation has two solutions for the time. Not all mathematical solutions make sense physically. Which solution makes sense physically in terms of the picture that you drew at the beginning? Keeping the same coordinate system, what is the velocity in the direction as a function of time? What is the velocity when the rock hits the bottom of the hole? ANSWER: Correct Part B How long is the rock in the air, from the instant it is released until it hits the bottom of the hole? Express your answer with the appropriate units. y = 0 m y t = 0 y y t y y v = -26.1 ms Hint 1. How to approach the problem How is the time the rock was in the air related to the time at which the rock hit the ground in Part A? ANSWER: Correct Enhanced EOC: Problem 2.23 A particle moving along the x-axis has its position described by the function 2.00 5.00 5.00 , where is in s. At = 4.00, what are the particle’s (a) position, (b) velocity, and (c) acceleration? You may want to review ( pages 38 – 42) . For help with math skills, you may want to review: Differentiation of Polynomial Functions t = 4.90 s x = ( t3 − t + ) m t t Part A Express your answer with the appropriate units. Hint 1. How to approach the problem Evaluate the position at time = 4.00 . ANSWER: Correct Part B Express your answer with the appropriate units. Hint 1. How to approach the problem How do you determine the velocity as a function of time, , from the position, ? What calculus operation do you have to perform? Once you have , how do you determine at a particular time? ANSWER: Correct Part C Express your answer with the appropriate units. t s 113 m v(t) x(t) v(t) v 91.0 ms Hint 1. How to approach the problem How do you determine the acceleration as a function of time, , from the velocity, ? What calculus operation do you have to perform? Once you have , how do you determine the acceleration at a particular time? ANSWER: Correct Problem 2.26 A particle’s position on the x-axis is given by the function 6.00 6.00 , where is in s. Part A Where is the particle when = 4.00 ? Express your answer with the appropriate units. ANSWER: Correct Problem 2.30 A particle’s velocity is described by the function = , where is in . a(t) v(t) a(t) 48.0 m s2 x = (t2 − t + ) m t vx m/s 1.00 m vx t2 − 7t + 7 m/s t s Part A How many turning points does the particle reach. Express your answer as an integer. ANSWER: Correct Part B At what times does the particle reach its turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct Part C What is the particle’s acceleration at each of the turning points? Express your answers using two significant figures separated by a comma. ANSWER: Correct 2 t1 , t2 = 5.8,1.2 s a1 , a2 = 4.6,-4.6 m/s2 Problem 2.49 A 200 weather rocket is loaded with 100 of fuel and fired straight up. It accelerates upward at 35 for 30 , then runs out of fuel. Ignore any air resistance effects. Part A What is the rocket’s maximum altitude? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long is the rocket in the air? Express your answer to two significant figures and include the appropriate units. ANSWER: Answer Requested Problem 2.52 A hotel elevator ascends with maximum speed of . Its acceleration and deceleration both have a magnitude of . Part A How far does the elevator move while accelerating to full speed from rest? kg kg m/s2 s h = 72 km t = 260 s 200 m 5 m/s 1.0 m/s2 Express your answer with the appropriate units. ANSWER: Correct Part B How long does it take to make the complete trip from bottom to top? Express your answer with the appropriate units. ANSWER: Answer Requested Components of Vectors Shown is a 10 by 10 grid, with coordinate axes x and y . The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through . This problem will ask you various questions about these vectors. All answers should be in decimal notation, unless otherwise specified. 12.5 m 45.0 s A D Part A What is the x component of ? Express your answer to two significant figures. Hint 1. How to derive the component A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You are asked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin (because the tail of the vector is at the origin). ANSWER: Correct Part B What is the y component of ? Express your answer to the nearest integer. ANSWER: Correct A A A A Ax = 2.5 A Ay = 3 Part C What is the y component of ? Express your answer to the nearest integer. Hint 1. Consider the direction Don’t forget the sign. ANSWER: Correct Part D What is the component of ? Express your answer to the nearest integer. Hint 1. How to find the start and end points of the vector components A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you need to somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow) of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the components of and . This is equivalent to the previous method, but it might be easier to visualize. ANSWER: B By = -3 x C C C A B Cx = -2 Correct The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first, followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation. The answers below are all integers, so estimate the components to the nearest whole number. Part E In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part F In ordered pair notation, write down the components of vector . Express your answers to the nearest integer. ANSWER: Correct Part G What is true about and ? Choose from the pulldown list below. A B Bx, By = 2,-3 D Dx, Dy = 2,-3 B D ANSWER: Correct Problem 3.6 Find x- and y-components of the following vectors. Part A Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Part B Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: They have different components and are not the same vectors. They have the same components but are not the same vectors. They are the same vectors. = (r 430m, 60& below positive x − axis) rx, ry = 210,-370 m v = (610m/s, 23& above positive x − axis) Correct Part C Express your answers using two significant figures. Enter your answers numerically separated by a comma. ANSWER: Correct Problem 3.10 Part A Draw . Draw the vector with its tail at the origin. ANSWER: vx, vy = 560,240 m/s a = (7.3m/s2 , negative y − direction) ax, ay = 0,-7.3 m/s2 B = −4 + 4 ı ^  ^ Correct Part B Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct B B = 5.7 Part C Find the direction of . Express your answer using two significant figures. ANSWER: Correct Part D Draw . Draw the vector with its tail at the origin. ANSWER: B = 45 above the B negative x-axis & = (−2.0 − 1.0 ) cm r ı ^  ^ Correct Part E Find the magnitude of . Express your answer using two significant figures. ANSWER: Correct r r = 2.2 cm Part F Find the direction of . ANSWER: Correct Part G Draw . Draw the vector with its tail at the origin. ANSWER: r = 26.6 below the r negative x-axis & = (−10 − 100 ) m/s v ı ^  ^ Correct Part H Find the magnitude of . Express your answer using four significant figures. ANSWER: Correct v v = 100.5 m/s Part I Find the direction of . ANSWER: Correct Part J Draw . Draw the vector with it’s tail at the origin. ANSWER: v = 84.3 below the v negative x-axis & = (20 + 10 ) m/ a ı ^  ^ s2 Correct Part K Find the magnitude of . ANSWER: Correct Part L a a = 22.4 m/s2 Find the direction of . ANSWER: Correct Problem 3.14 Let , , and . Part A What is the component form of vector ? ANSWER: Correct Part B What is the magnitude of vector ? ANSWER: a = 26.6 above the a positive x-axis & A = 5 − 2 ı ^  ^ B = −2 + 6 ı ^  ^ D = A − B D D = 7 − 8 ı ^  ^ D = −7 − 5 ı ^  ^ D = 7 + 8 ı ^  ^ D = 4 + 5 ı ^  ^ D Correct Part C What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.15 Let , , and . Part A Write vector in component form. ANSWER: D = 10.6 D  = 49 & below positive x-axis A = 4 − 2 ı ^  ^ B = −3 + 5 ı ^  ^ E = 4A + 2B E E = 10 + 2 ı ^  ^ E = + 10 ı ^  ^ E = −10 ^ E = 10 − 2 ı ^  ^ Correct Part B Draw vectors , , and . Draw the vectors with their tails at the origin. ANSWER: Correct Part C A B E What is the magnitude of vector ? Express your answer using two significant figures. ANSWER: Correct Part D What is the direction of vector ? Express your answer using two significant figures. ANSWER: Correct Problem 3.24 Part A What is the angle between vectors and in the figure? Express your answer with the appropriate units. E E = 10.0 E  = 11 & counterclockwise from positive direction of x-axis  E F ANSWER: Correct Part B Use components to determine the magnitude of . ANSWER: Correct Part C Use components to determine the direction of . Express your answer with the appropriate units. ANSWER: Correct Score Summary: Your score on this assignment is 91.3%.  = 71.6 & G = E + F  G = 3.00 G = E + F   = 90.0 & You received 129.62 out of a possible total of 142 points.

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In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

In case the body have to stay in lower temperature for extended time period (more than 1 hour), how does the body regulate its response?

Arterioles transporting blood to external capillaries beneath the surface of … Read More...
Assignment 10 Due: 11:59pm on Wednesday, April 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Ka Kc Correct Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: I2 I1 Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct I2/I1 = 32 rpm s  = 440 rad s2 s  = 12 rev Constant Angular Acceleration in the Kitchen Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. Hint 1. How to approach the problem Recall from your study of kinematics the three equations of motion derived for systems undergoing constant linear acceleration. You are now studying systems undergoing constant angular acceleration and will need to work with the three analogous equations of motion. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find the angular acceleration . Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it What is the angular velocity of the salad spinner as Dario is spinning it? Express your answer numerically in degrees per second. Hint 1. Converting rotations to degrees When the salad spinner spins through one revolution, it turns through 360 degrees. ANSWER: Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many degrees does the salad spinner rotate as it comes to rest? Express your answer numerically in degrees. Hint 1. Converting rotations to degrees  0 = 1440 degrees/s  =  − 0 One revolution is equivalent to 360 degrees. ANSWER: Hint 4. Determine which equation to use You know the initial and final velocities of the system and the angular distance through which the spinner rotates as it comes to a stop. Which equation should be used to solve for the unknown constant angular acceleration ? ANSWER: ANSWER: Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds.  = 2160 degrees   = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0  = -480 degrees/s2 Hint 1. How to approach the problem Again, you will need the equations of rotational kinematics that apply to situations of constant angular acceleration. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find . Hint 2. Determine which equation to use You have the initial and final velocities of the system and the angular acceleration, which you found in the previous part. Which is the best equation to use to solve for the unknown time ? ANSWER: ANSWER: Correct ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. Hint 1. Average acceleration Recall that if the angular velocity decreases uniformly, the angular acceleration will remain constant. Therefore, the angular acceleration is just the total change in angular velocity divided by t t  = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0 t = 3.00 s rev/min rev/min s  the total change in time. Be careful of the sign of the angular acceleration. ANSWER: Correct Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. Hint 1. Determine the correct kinematic equation Which of the following kinematic equations is best suited to this problem? Here and are the initial and final angular velocities, is the elapsed time, is the constant angular acceleration, and and are the initial and final angular displacements. Hint 1. How to chose the right equation Notice that you were given in the problem introduction the initial and final speeds, as well as the length of time between them. In this problem, you are asked to find the number of revolutions (which here is the change in angular displacement, ). If you already found the angular acceleration in Part A, you could use that as well, but you would end up using a more complex equation. Also, in general, it is somewhat favorable to use given quantities instead of quantities that you have calculated. ANSWER:  = -1.10 rev/s2 0  t  0   − 0  = 0 + t  = 0 + t+  1 2 t2 = + 2( − ) 2 20 0 − 0 = (+ )t 1 2 0 ANSWER: Correct Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? Hint 1. Finding the total time for spin down To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for the time. One can then just subtract the time it took to reach 250 from the total time. Be careful of your signs when you set up the equation. ANSWER: Correct Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: 29.0 rev rev/min 3.79 s g g cm rpm g Correct Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? Express your answer with the appropriate units. ANSWER: Correct Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: v = 3.2 ms g cm J 3.65 ms cm g rpm K = 4.4 J Correct Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. Hint 1. Conditions for equilibrium If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case in the current situation where block B has twice the mass of block A? Hint 2. Rotational analogue of Newton’s second law The net torque of all the forces acting on a rigid body is proportional to the angular acceleration of the body net  and is given by , where is the moment of inertia of the body. Hint 3. Relation between linear and angular acceleration A particle that rotates with angular acceleration has linear acceleration equal to , where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius of the pulley. ANSWER: Correct Part B What happens when block B moves downward? Hint 1. How to approach the problem To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley’s angular acceleration. Hint 2. Find the net torque on the pulley Let’s assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley? Take the positive sense of rotation to be counterclockwise. Express your answer in terms of , , and . net = I I  a a = R R a  = a R R The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. T1 T2 R net T1 T2 R Hint 1. Torque The torque of a force with respect to a point is defined as the product of the magnitude times the perpendicular distance between the line of action of and the point . In other words, . ANSWER: ANSWER: Correct Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration. Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units.  F  O F l F  O  = Fl net = R(T2 − T1 ) The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord. ANSWER: Correct Part B What is the direction of net torque about the axle? ANSWER: Correct Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units.  = 0.20 Nm Clockwise Counterclockwise kg cm kg ANSWER: Correct Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod.  = 41 Nm 45  = 29 Nm Icm Ip Ip = Icm + Md2 d M L m Icm = m 1 12 L2 Iend Iend m L d ANSWER: ANSWER: Correct Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the axis Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = L 2 Iend = mL2 3 m a Icm Icm = m 1 6 a2 Iedge Iedge m a o p d ANSWER: Correct Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: d = a 2 Iedge = 2ma2 3 cm s rpm 2.5 × 10−5 kg m2 = 1.8×10−3  Nm Correct Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Correct Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. N = 48 rev kgm2 rad/s2 8.14 N  m kg kg m Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A kg d = 1.4 m M1 = 3.6 kg L = 0.66 m L1 = 0.33 m L2 = 0.15 m  = 17 g = 9.8 m/s2 Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. Hint 1. Nature of the problem Remember that this is a statics problem, so all forces and torques are balanced (their sums equal zero). Hint 2. Origin of torque Calculate the torque about the point at which the arm attaches to the rest of the body. This allows one to balance the torques without having to worry about the undefined forces at this point. Hint 3. Adding up the torques Add up the torques about the point in which the humerus attaches to the body. Answer in terms of , , , , , and . Remember that counterclockwise torque is positive. ANSWER: ANSWER: Correct Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. T L1 L2 M1 g T  total = 0 = L1M1g − Tsin()L2 T = 265 N Fy Hint 1. Total forces involved Recall that there are three vertical forces in this problem: the force of gravity acting on the bone, the force from the vertical component of the muscle tension, and the force exerted by the scapula on the humerus (where it attaches to the rest of the body). ANSWER: Correct Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: Correct ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). |Fy| = 42 N Fx |Fx| = 254 N F 1 F  2 F  3 Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. Hint 1. Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? Give your answer numerically in newton-meters to two significant figures. ANSWER: ANSWER: Correct Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. F3 = 0 F1 = 12 N F 2 F 1   1 F  1 |  1 | = 0.36 N  m F2 = 4.5 N I = 9 kg m2 F1 = 12 N F2 = 27 N F3 = 0 t 45  Hint 1. Find the net torque about the pivot What is the magnitude of the total torque around the pivot point? Answer numerically in newton-meters to two significant figures. ANSWER: Hint 2. Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Express your answer numerically in radians per second squared to two significant figures. Hint 1. Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from ANSWER: Hint 3. Description of angular kinematics Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go through a given angle . For constant acceleration ( ) and starting with (where is angular speed) the relation is given by which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, | p ivot| | p ivot| = 1.8 N  m    vot Ivot  pivot = Ipivot.  = 0.20 radians/s2    = 0   = 1  , 2 t2 x a . ANSWER: Correct Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. Hint 1. Find the required component of Only the tangential (perpendicular) component of (call it ) provides a torque. What is ? Answer in terms of . You will need to evaluate any trigonometric functions. ANSWER: ANSWER: Correct x = 1 a 2 t2 t = 2.8 s F1 = 12 N F2 = 0 F3 F3 F 3 F  3 F3t F3t F3 F3t = 1 2 F3 F3 = 9.0 N Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: cm m/s 833 rpm 48.0 ms 0 ms Correct Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. ANSWER: Correct Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? g m/s 0.583 J g ANSWER: Correct Part B What is the direction of the angular momentum of the bar ? ANSWER: Correct Problem 12.46 Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? 3.27 kgm2/s into the page out of the page kg ANSWER: Correct Part B What is its direction? ANSWER: Correct Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. 3.66×10−2 kgm /s 2 x direction -x direction y direction -y direction z direction -z direction m Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1.  = 47 m kg kg m Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: F1 = 670 N F2 = 900 N kg m kg The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture of the four forces acting on the beam, indicating both their direction and the place on the beam that the forces are acting. Choose a coordinate system with a direction for the axis along the beam, and indicate the position of the boy. What is the net force on the beam if it is stationary? Just before the beam tips, the force of the left support on the beam is zero. Using the zero net force condition, what is the force due to the right support just before the beam tips? For the beam to remain stationary, what must be zero besides the net force on the beam? Choose a point on the beam, and compute the net torque on the beam about that point. Be sure to choose a positive direction for the rotation axis and therefore the torques. Using the zero torque condition, what is the position of the boy on the beam just prior to tipping? How far is this position from the right edge of the beam? ANSWER: Correct d = 2.0 m Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: m kg rpm N  m t = 250 s = 1.1×106 E J s Correct Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? P = 2.4×105 W  = 1800 Nm kg rpm s Express your answer with the appropriate units. ANSWER: Correct Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: 5.76 N kg cm  = 22 rad s2 Correct Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Score Summary:  = 6.6 rad s kg rev/s kg cm cm kg cm kg cm cm 2 = Your score on this assignment is 95.7%. You received 189.42 out of a possible total of 198 points.

Assignment 10 Due: 11:59pm on Wednesday, April 23, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 12.3 Part A The figure shows three rotating disks, all of equal mass. Rank in order, from largest to smallest, their rotational kinetic energies to . Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: Ka Kc Correct Conceptual Question 12.6 You have two steel solid spheres. Sphere 2 has twice the radius of sphere 1. Part A By what factor does the moment of inertia of sphere 2 exceed the moment of inertia of sphere 1? ANSWER: I2 I1 Correct Problem 12.2 A high-speed drill reaches 2500 in 0.59 . Part A What is the drill’s angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B Through how many revolutions does it turn during this first 0.59 ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct I2/I1 = 32 rpm s  = 440 rad s2 s  = 12 rev Constant Angular Acceleration in the Kitchen Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. Part A What is the angular acceleration of the salad spinner as it slows down? Express your answer numerically in degrees per second per second. Hint 1. How to approach the problem Recall from your study of kinematics the three equations of motion derived for systems undergoing constant linear acceleration. You are now studying systems undergoing constant angular acceleration and will need to work with the three analogous equations of motion. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find the angular acceleration . Hint 2. Find the angular velocity of the salad spinner while Dario is spinning it What is the angular velocity of the salad spinner as Dario is spinning it? Express your answer numerically in degrees per second. Hint 1. Converting rotations to degrees When the salad spinner spins through one revolution, it turns through 360 degrees. ANSWER: Hint 3. Find the angular distance the salad spinner travels as it comes to rest Through how many degrees does the salad spinner rotate as it comes to rest? Express your answer numerically in degrees. Hint 1. Converting rotations to degrees  0 = 1440 degrees/s  =  − 0 One revolution is equivalent to 360 degrees. ANSWER: Hint 4. Determine which equation to use You know the initial and final velocities of the system and the angular distance through which the spinner rotates as it comes to a stop. Which equation should be used to solve for the unknown constant angular acceleration ? ANSWER: ANSWER: Correct Part B How long does it take for the salad spinner to come to rest? Express your answer numerically in seconds.  = 2160 degrees   = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0  = -480 degrees/s2 Hint 1. How to approach the problem Again, you will need the equations of rotational kinematics that apply to situations of constant angular acceleration. Collect your known quantities and then determine which of the angular kinematic equations is appropriate to find . Hint 2. Determine which equation to use You have the initial and final velocities of the system and the angular acceleration, which you found in the previous part. Which is the best equation to use to solve for the unknown time ? ANSWER: ANSWER: Correct ± A Spinning Electric Fan An electric fan is turned off, and its angular velocity decreases uniformly from 540 to 250 in a time interval of length 4.40 . Part A Find the angular acceleration in revolutions per second per second. Hint 1. Average acceleration Recall that if the angular velocity decreases uniformly, the angular acceleration will remain constant. Therefore, the angular acceleration is just the total change in angular velocity divided by t t  = 0 + 0t+  1 2 t2  = 0 + t = + 2( − ) 2 20 0 t = 3.00 s rev/min rev/min s  the total change in time. Be careful of the sign of the angular acceleration. ANSWER: Correct Part B Find the number of revolutions made by the fan blades during the time that they are slowing down in Part A. Hint 1. Determine the correct kinematic equation Which of the following kinematic equations is best suited to this problem? Here and are the initial and final angular velocities, is the elapsed time, is the constant angular acceleration, and and are the initial and final angular displacements. Hint 1. How to chose the right equation Notice that you were given in the problem introduction the initial and final speeds, as well as the length of time between them. In this problem, you are asked to find the number of revolutions (which here is the change in angular displacement, ). If you already found the angular acceleration in Part A, you could use that as well, but you would end up using a more complex equation. Also, in general, it is somewhat favorable to use given quantities instead of quantities that you have calculated. ANSWER:  = -1.10 rev/s2 0  t  0   − 0  = 0 + t  = 0 + t+  1 2 t2 = + 2( − ) 2 20 0 − 0 = (+ )t 1 2 0 ANSWER: Correct Part C How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in Part A? Hint 1. Finding the total time for spin down To find the total time for spin down, just calculate when the velocity will equal zero. This is accomplished by setting the initial velocity plus the acceleration multipled by the time equal to zero and then solving for the time. One can then just subtract the time it took to reach 250 from the total time. Be careful of your signs when you set up the equation. ANSWER: Correct Problem 12.8 A 100 ball and a 230 ball are connected by a 34- -long, massless, rigid rod. The balls rotate about their center of mass at 130 . Part A What is the speed of the 100 ball? Express your answer to two significant figures and include the appropriate units. ANSWER: 29.0 rev rev/min 3.79 s g g cm rpm g Correct Problem 12.10 A thin, 60.0 disk with a diameter of 9.00 rotates about an axis through its center with 0.200 of kinetic energy. Part A What is the speed of a point on the rim? Express your answer with the appropriate units. ANSWER: Correct Problem 12.12 A drum major twirls a 95- -long, 470 baton about its center of mass at 150 . Part A What is the baton’s rotational kinetic energy? Express your answer to two significant figures and include the appropriate units. ANSWER: v = 3.2 ms g cm J 3.65 ms cm g rpm K = 4.4 J Correct Net Torque on a Pulley The figure below shows two blocks suspended by a cord over a pulley. The mass of block B is twice the mass of block A, while the mass of the pulley is equal to the mass of block A. The blocks are let free to move and the cord moves on the pulley without slipping or stretching. There is no friction in the pulley axle, and the cord’s weight can be ignored. Part A Which of the following statements correctly describes the system shown in the figure? Check all that apply. Hint 1. Conditions for equilibrium If the blocks had the same mass, the system would be in equilibrium. The blocks would have zero acceleration and the tension in each part of the cord would equal the weight of each block. Both parts of the cord would then pull with equal force on the pulley, resulting in a zero net torque and no rotation of the pulley. Is this still the case in the current situation where block B has twice the mass of block A? Hint 2. Rotational analogue of Newton’s second law The net torque of all the forces acting on a rigid body is proportional to the angular acceleration of the body net  and is given by , where is the moment of inertia of the body. Hint 3. Relation between linear and angular acceleration A particle that rotates with angular acceleration has linear acceleration equal to , where is the distance of the particle from the axis of rotation. In the present case, where there is no slipping or stretching of the cord, the cord and the pulley must move together at the same speed. Therefore, if the cord moves with linear acceleration , the pulley must rotate with angular acceleration , where is the radius of the pulley. ANSWER: Correct Part B What happens when block B moves downward? Hint 1. How to approach the problem To determine whether the tensions in both parts of the cord are equal, it is convenient to write a mathematical expression for the net torque on the pulley. This will allow you to relate the tensions in the cord to the pulley’s angular acceleration. Hint 2. Find the net torque on the pulley Let’s assume that the tensions in both parts of the cord are different. Let be the tension in the right cord and the tension in the left cord. If is the radius of the pulley, what is the net torque acting on the pulley? Take the positive sense of rotation to be counterclockwise. Express your answer in terms of , , and . net = I I  a a = R R a  = a R R The acceleration of the blocks is zero. The net torque on the pulley is zero. The angular acceleration of the pulley is nonzero. T1 T2 R net T1 T2 R Hint 1. Torque The torque of a force with respect to a point is defined as the product of the magnitude times the perpendicular distance between the line of action of and the point . In other words, . ANSWER: ANSWER: Correct Note that if the pulley were stationary (as in many systems where only linear motion is studied), then the tensions in both parts of the cord would be equal. However, if the pulley rotates with a certain angular acceleration, as in the present situation, the tensions must be different. If they were equal, the pulley could not have an angular acceleration. Problem 12.18 Part A In the figure , what is the magnitude of net torque about the axle? Express your answer to two significant figures and include the appropriate units.  F  O F l F  O  = Fl net = R(T2 − T1 ) The left cord pulls on the pulley with greater force than the right cord. The left and right cord pull with equal force on the pulley. The right cord pulls on the pulley with greater force than the left cord. ANSWER: Correct Part B What is the direction of net torque about the axle? ANSWER: Correct Problem 12.22 An athlete at the gym holds a 3.5 steel ball in his hand. His arm is 78 long and has a mass of 3.6 . Assume the center of mass of the arm is at the geometrical center of the arm. Part A What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Express your answer to two significant figures and include the appropriate units.  = 0.20 Nm Clockwise Counterclockwise kg cm kg ANSWER: Correct Part B What is the magnitude of the torque about his shoulder if he holds his arm straight, but below horizontal? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Parallel Axis Theorem The parallel axis theorem relates , the moment of inertia of an object about an axis passing through its center of mass, to , the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is , where is the perpendicular distance from the center of mass to the axis that passes through point p, and is the mass of the object. Part A Suppose a uniform slender rod has length and mass . The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by . Find , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the center of mass Find the distance appropriate to this problem. That is, find the perpendicular distance from the center of mass of the rod to the axis passing through one end of the rod.  = 41 Nm 45  = 29 Nm Icm Ip Ip = Icm + Md2 d M L m Icm = m 1 12 L2 Iend Iend m L d ANSWER: ANSWER: Correct Part B Now consider a cube of mass with edges of length . The moment of inertia of the cube about an axis through its center of mass and perpendicular to one of its faces is given by . Find , the moment of inertia about an axis p through one of the edges of the cube Express in terms of and . Use fractions rather than decimal numbers in your answer. Hint 1. Find the distance from the axis to the axis Find the perpendicular distance from the center of mass axis to the new edge axis (axis labeled p in the figure). ANSWER: d = L 2 Iend = mL2 3 m a Icm Icm = m 1 6 a2 Iedge Iedge m a o p d ANSWER: Correct Problem 12.26 Starting from rest, a 12- -diameter compact disk takes 2.9 to reach its operating angular velocity of 2000 . Assume that the angular acceleration is constant. The disk’s moment of inertia is . Part A How much torque is applied to the disk? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How many revolutions does it make before reaching full speed? Express your answer using two significant figures. ANSWER: d = a 2 Iedge = 2ma2 3 cm s rpm 2.5 × 10−5 kg m2 = 1.8×10−3  Nm Correct Problem 12.23 An object’s moment of inertia is 2.20 . Its angular velocity is increasing at the rate of 3.70 . Part A What is the total torque on the object? ANSWER: Correct Problem 12.31 A 5.1 cat and a 2.5 bowl of tuna fish are at opposite ends of the 4.0- -long seesaw. N = 48 rev kgm2 rad/s2 8.14 N  m kg kg m Part A How far to the left of the pivot must a 3.8 cat stand to keep the seesaw balanced? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Static Equilibrium of the Arm You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass , length and its center of mass is a distance from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance from the scapula. The deltoid muscle makes an angle of with the horizontal, as shown. Use throughout the problem. Part A kg d = 1.4 m M1 = 3.6 kg L = 0.66 m L1 = 0.33 m L2 = 0.15 m  = 17 g = 9.8 m/s2 Find the tension in the deltoid muscle. Express the tension in newtons, to the nearest integer. Hint 1. Nature of the problem Remember that this is a statics problem, so all forces and torques are balanced (their sums equal zero). Hint 2. Origin of torque Calculate the torque about the point at which the arm attaches to the rest of the body. This allows one to balance the torques without having to worry about the undefined forces at this point. Hint 3. Adding up the torques Add up the torques about the point in which the humerus attaches to the body. Answer in terms of , , , , , and . Remember that counterclockwise torque is positive. ANSWER: ANSWER: Correct Part B Using the conditions for static equilibrium, find the magnitude of the vertical component of the force exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Express your answer in newtons, to the nearest integer. T L1 L2 M1 g T  total = 0 = L1M1g − Tsin()L2 T = 265 N Fy Hint 1. Total forces involved Recall that there are three vertical forces in this problem: the force of gravity acting on the bone, the force from the vertical component of the muscle tension, and the force exerted by the scapula on the humerus (where it attaches to the rest of the body). ANSWER: Correct Part C Now find the magnitude of the horizontal component of the force exerted by the scapula on the humerus. Express your answer in newtons, to the nearest integer. ANSWER: Correct ± Moments around a Rod A rod is bent into an L shape and attached at one point to a pivot. The rod sits on a frictionless table and the diagram is a view from above. This means that gravity can be ignored for this problem. There are three forces that are applied to the rod at different points and angles: , , and . Note that the dimensions of the bent rod are in centimeters in the figure, although the answers are requested in SI units (kilograms, meters, seconds). |Fy| = 42 N Fx |Fx| = 254 N F 1 F  2 F  3 Part A If and , what does the magnitude of have to be for there to be rotational equilibrium? Answer numerically in newtons to two significant figures. Hint 1. Finding torque about pivot from What is the magnitude of the torque | | provided by around the pivot point? Give your answer numerically in newton-meters to two significant figures. ANSWER: ANSWER: Correct Part B If the L-shaped rod has a moment of inertia , , , and again , how long a time would it take for the object to move through ( /4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures. F3 = 0 F1 = 12 N F 2 F 1   1 F  1 |  1 | = 0.36 N  m F2 = 4.5 N I = 9 kg m2 F1 = 12 N F2 = 27 N F3 = 0 t 45  Hint 1. Find the net torque about the pivot What is the magnitude of the total torque around the pivot point? Answer numerically in newton-meters to two significant figures. ANSWER: Hint 2. Calculate Given the total torque around the pivot point, what is , the magnitude of the angular acceleration? Express your answer numerically in radians per second squared to two significant figures. Hint 1. Equation for If you know the magnitude of the total torque ( ) and the rotational inertia ( ), you can then find the rotational acceleration ( ) from ANSWER: Hint 3. Description of angular kinematics Now that you know the angular acceleration, this is a problem in rotational kinematics; find the time needed to go through a given angle . For constant acceleration ( ) and starting with (where is angular speed) the relation is given by which is analogous to the expression for linear displacement ( ) with constant acceleration ( ) starting from rest, | p ivot| | p ivot| = 1.8 N  m    vot Ivot  pivot = Ipivot.  = 0.20 radians/s2    = 0   = 1  , 2 t2 x a . ANSWER: Correct Part C Now consider the situation in which and , but now a force with nonzero magnitude is acting on the rod. What does have to be to obtain equilibrium? Give a numerical answer, without trigonometric functions, in newtons, to two significant figures. Hint 1. Find the required component of Only the tangential (perpendicular) component of (call it ) provides a torque. What is ? Answer in terms of . You will need to evaluate any trigonometric functions. ANSWER: ANSWER: Correct x = 1 a 2 t2 t = 2.8 s F1 = 12 N F2 = 0 F3 F3 F 3 F  3 F3t F3t F3 F3t = 1 2 F3 F3 = 9.0 N Problem 12.32 A car tire is 55.0 in diameter. The car is traveling at a speed of 24.0 . Part A What is the tire’s rotation frequency, in rpm? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part B What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropriate units. ANSWER: Correct Part C What is the speed of a point at the bottom edge of the tire? Express your answer as an integer and include the appropriate units. ANSWER: cm m/s 833 rpm 48.0 ms 0 ms Correct Problem 12.33 A 460 , 8.00-cm-diameter solid cylinder rolls across the floor at 1.30 . Part A What is the can’s kinetic energy? Express your answer with the appropriate units. ANSWER: Correct Problem 12.45 Part A What is the magnitude of the angular momentum of the 780 rotating bar in the figure ? g m/s 0.583 J g ANSWER: Correct Part B What is the direction of the angular momentum of the bar ? ANSWER: Correct Problem 12.46 Part A What is the magnitude of the angular momentum of the 2.20 , 4.60-cm-diameter rotating disk in the figure ? 3.27 kgm2/s into the page out of the page kg ANSWER: Correct Part B What is its direction? ANSWER: Correct Problem 12.60 A 3.0- -long ladder, as shown in the following figure, leans against a frictionless wall. The coefficient of static friction between the ladder and the floor is 0.46. 3.66×10−2 kgm /s 2 x direction -x direction y direction -y direction z direction -z direction m Part A What is the minimum angle the ladder can make with the floor without slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.61 The 3.0- -long, 90 rigid beam in the following figure is supported at each end. An 70 student stands 2.0 from support 1.  = 47 m kg kg m Part A How much upward force does the support 1 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much upward force does the support 2 exert on the beam? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Enhanced EOC: Problem 12.63 A 44 , 5.5- -long beam is supported, but not attached to, the two posts in the figure . A 22 boy starts walking along the beam. You may want to review ( pages 330 – 334) . For help with math skills, you may want to review: F1 = 670 N F2 = 900 N kg m kg The Vector Cross Product Part A How close can he get to the right end of the beam without it falling over? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem Draw a picture of the four forces acting on the beam, indicating both their direction and the place on the beam that the forces are acting. Choose a coordinate system with a direction for the axis along the beam, and indicate the position of the boy. What is the net force on the beam if it is stationary? Just before the beam tips, the force of the left support on the beam is zero. Using the zero net force condition, what is the force due to the right support just before the beam tips? For the beam to remain stationary, what must be zero besides the net force on the beam? Choose a point on the beam, and compute the net torque on the beam about that point. Be sure to choose a positive direction for the rotation axis and therefore the torques. Using the zero torque condition, what is the position of the boy on the beam just prior to tipping? How far is this position from the right edge of the beam? ANSWER: Correct d = 2.0 m Problem 12.68 Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel’s energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.6 diameter and a mass of 270 . Its maximum angular velocity is 1500 . Part A A motor spins up the flywheel with a constant torque of 54 . How long does it take the flywheel to reach top speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How much energy is stored in the flywheel? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C The flywheel is disconnected from the motor and connected to a machine to which it will deliver energy. Half the energy stored in the flywheel is delivered in 2.2 . What is the average power delivered to the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: m kg rpm N  m t = 250 s = 1.1×106 E J s Correct Part D How much torque does the flywheel exert on the machine? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.71 The 3.30 , 40.0-cm-diameter disk in the figure is spinning at 350 . Part A How much friction force must the brake apply to the rim to bring the disk to a halt in 2.10 ? P = 2.4×105 W  = 1800 Nm kg rpm s Express your answer with the appropriate units. ANSWER: Correct Problem 12.74 A 5.0 , 60- -diameter cylinder rotates on an axle passing through one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. Part A What is the magnitude of the cylinder’s initial angular acceleration? Express your answer to two significant figures and include the appropriate units. ANSWER: 5.76 N kg cm  = 22 rad s2 Correct Part B What is the magnitude of the cylinder’s angular velocity when it is directly below the axle? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 12.82 A 45 figure skater is spinning on the toes of her skates at 0.90 . Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 , 20 average diameter, 160 tall) plus two rod-like arms (2.5 each, 67 long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 , 20- -diameter, 200- -tall cylinder. Part A What is her new rotation frequency, in revolutions per second? Express your answer to two significant figures and include the appropriate units. ANSWER: Incorrect; Try Again Score Summary:  = 6.6 rad s kg rev/s kg cm cm kg cm kg cm cm 2 = Your score on this assignment is 95.7%. You received 189.42 out of a possible total of 198 points.

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Problems Marking scheme 1. Let A be a nonzero square matrix. Is it possible that a positive integer k exists such that ?? = 0 ? For example, find ?3 for the matrix [ 0 1 2 0 0 1 0 0 0 ] A square matrix A is nilpotent of index k when ? ≠ 0 , ?2 ≠ 0 , … . . , ??−1 ≠ 0, ??? ?? = 0. In this task you will explore nilpotent matrices. 1. The matrix in the example given above is nilpotent. What is its index? ( 2 marks ) 2. Use a software program to determine which of the following matrices are nilpotent and find their indices ( 12 marks ) A. [ 0 1 0 0 ] B. [ 0 1 1 0 ] C. [ 0 0 1 0 ] D. [ 1 0 1 0 ] E. [ 0 0 1 0 0 0 0 0 0 ] F. [ 0 0 0 1 0 0 1 1 0 ] 3. Find 3×3 nilpotent matrices of indices 2 and 3 ( 2 marks ) 4. Find 4×4 nilpotent matrices of indices 2, 3, and 4 ( 2 marks ) 5. Find nilpotent matrix of index 5 ( 2 marks ) 6. Are nilpotent matrices invertible? prove your answer ( 3 marks ) 7. When A is nilpotent, what can you say about ?? ? prove your answer ( 3 marks ) 8. Show that if ? is nilpotent , then ? − ? is invertible ( 4 marks ) 30% 2. A radio transmitter circuit contains a resisitance of 2.0 Ω, a variable inductor of 100 − ? ℎ?????? and a voltage source of 4.0 ? . find the current ? in the circuit as a function of the time t for 0 ≤ ? ≤ 100? if the intial curent is zero. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 3. An object falling under the influence of gravity has a variable accelertaion given by 32 − ? , where ? represents the velocity. If the object starts from rest, find an expression for the velocity in terms of the time. Also, find the limiting value of the velocity. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 4. When the angular displacement ? of a pendulum is small ( less than 60), the pendulum moves with simple harmonic motion closely approximated by ?′′ + ? ? ? = 0 . Here , ?′ = ?? ?? and ? is the accelertaion due to gravity , and ? is the length of the pendulum. Find ? as a function of time ( in s ) if ? = 9.8 ?/?2, ? = 1.0 ? ? = 0.1 and ?? ?? = 0 when ? = 0 . sketch the cuve using any graphical tool. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 5. Find the equation relating the charge and the time in a electric circuit with the following elements: ? = 0.200 ? , ? = 8.00 Ω , ? = 1.00 ?? , ? = 0. In this circuit , ? = 0 and ? = 0.500 ? when ? = 0 Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 6. A spring is stretched 1 m by ? 20 − ? Weight. The spring is stretched 0.5 m below the equilibrium position with the weight attached and the then released. If it is a medium that resists the motion with a force equal to 12?, where v is the velocity, sketch and find the displacement y of the weight as a function of the time. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 7. A 20?? inductor, a 40.0 Ω resistor, a 50.0 ?? capacitor, and voltage source of 100 ?−100?are connected in series in an electric circuit. Find the charge on the capacitor as a function of time t , if ? = 0 and ? = 0 ?ℎ?? ? = 0 Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 10% quality and neatness and using Math equations in MS word. –

Problems Marking scheme 1. Let A be a nonzero square matrix. Is it possible that a positive integer k exists such that ?? = 0 ? For example, find ?3 for the matrix [ 0 1 2 0 0 1 0 0 0 ] A square matrix A is nilpotent of index k when ? ≠ 0 , ?2 ≠ 0 , … . . , ??−1 ≠ 0, ??? ?? = 0. In this task you will explore nilpotent matrices. 1. The matrix in the example given above is nilpotent. What is its index? ( 2 marks ) 2. Use a software program to determine which of the following matrices are nilpotent and find their indices ( 12 marks ) A. [ 0 1 0 0 ] B. [ 0 1 1 0 ] C. [ 0 0 1 0 ] D. [ 1 0 1 0 ] E. [ 0 0 1 0 0 0 0 0 0 ] F. [ 0 0 0 1 0 0 1 1 0 ] 3. Find 3×3 nilpotent matrices of indices 2 and 3 ( 2 marks ) 4. Find 4×4 nilpotent matrices of indices 2, 3, and 4 ( 2 marks ) 5. Find nilpotent matrix of index 5 ( 2 marks ) 6. Are nilpotent matrices invertible? prove your answer ( 3 marks ) 7. When A is nilpotent, what can you say about ?? ? prove your answer ( 3 marks ) 8. Show that if ? is nilpotent , then ? − ? is invertible ( 4 marks ) 30% 2. A radio transmitter circuit contains a resisitance of 2.0 Ω, a variable inductor of 100 − ? ℎ?????? and a voltage source of 4.0 ? . find the current ? in the circuit as a function of the time t for 0 ≤ ? ≤ 100? if the intial curent is zero. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 3. An object falling under the influence of gravity has a variable accelertaion given by 32 − ? , where ? represents the velocity. If the object starts from rest, find an expression for the velocity in terms of the time. Also, find the limiting value of the velocity. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 4. When the angular displacement ? of a pendulum is small ( less than 60), the pendulum moves with simple harmonic motion closely approximated by ?′′ + ? ? ? = 0 . Here , ?′ = ?? ?? and ? is the accelertaion due to gravity , and ? is the length of the pendulum. Find ? as a function of time ( in s ) if ? = 9.8 ?/?2, ? = 1.0 ? ? = 0.1 and ?? ?? = 0 when ? = 0 . sketch the cuve using any graphical tool. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 5. Find the equation relating the charge and the time in a electric circuit with the following elements: ? = 0.200 ? , ? = 8.00 Ω , ? = 1.00 ?? , ? = 0. In this circuit , ? = 0 and ? = 0.500 ? when ? = 0 Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 6. A spring is stretched 1 m by ? 20 − ? Weight. The spring is stretched 0.5 m below the equilibrium position with the weight attached and the then released. If it is a medium that resists the motion with a force equal to 12?, where v is the velocity, sketch and find the displacement y of the weight as a function of the time. Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 7. A 20?? inductor, a 40.0 Ω resistor, a 50.0 ?? capacitor, and voltage source of 100 ?−100?are connected in series in an electric circuit. Find the charge on the capacitor as a function of time t , if ? = 0 and ? = 0 ?ℎ?? ? = 0 Correct solution 5% Graph the general solution 2.5% Graph the function and particular solution 2.5% 10% quality and neatness and using Math equations in MS word. –

Problems Marking scheme 1. Let A be a nonzero square … Read More...
Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

Assignment 9 Due: 11:59pm on Friday, April 11, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 11.2 Part A Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Part B Evaluate the dot product if and . Express your answer using two significant figures. ANSWER: Correct Problem 11.4  A B = 5 − 6 A i ^ j ^ = −9 − 5 B i ^ j ^ A  B  = -15  A B = −5 + 9 A i ^ j ^ = 5 + 6 B i ^ j ^ A  B  = 29 Part A What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B What is the angle between vectors and if and ? Express your answer as an integer and include the appropriate units. ANSWER: Correct ± All Work and No Play Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement If an object undergoes displacement while being acted upon by a force (or several forces), it is said that work is being done on the object. If the object is moving in a straight line and the displacement and the force are known, the work done by the force can be calculated as , where is the work done by force on the object that undergoes displacement directed at angle relative to .  A B A = 2 + 5 ı ^  ^ B = −2 − 4 ı ^  ^  = 175  A B A = −6 + 2 ı ^  ^ B = − − 3 ı ^  ^  = 90 W =  = cos  F  s  F   s  W F  s  F  Note that depending on the value of , the work done can be positive, negative, or zero. In this problem, you will practice calculating work done on an object moving in a straight line. The first series of questions is related to the accompanying figure. Part A What can be said about the sign of the work done by the force ? ANSWER: Correct When , the cosine of is zero, and therefore the work done is zero. Part B cos  F  1 It is positive. It is negative. It is zero. There is not enough information to answer the question.  = 90  What can be said about the work done by force ? ANSWER: Correct When , is positive, and so the work done is positive. Part C The work done by force is ANSWER: Correct When , is negative, and so the work done is negative. Part D The work done by force is ANSWER: F  2 It is positive. It is negative. It is zero. 0 <  < 90 cos  F  3 positive negative zero 90 <  < 180 cos  F  4 Correct Part E The work done by force is ANSWER: Correct positive negative zero F  5 positive negative zero Part F The work done by force is ANSWER: Correct Part G The work done by force is ANSWER: Correct In the next series of questions, you will use the formula to calculate the work done by various forces on an object that moves 160 meters to the right. F  6 positive negative zero F  7 positive negative zero W =  = cos  F  s  F   s  Part H Find the work done by the 18-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part I Find the work done by the 30-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Part J Find the work done by the 12-newton force. Use two significant figures in your answer. Express your answer in joules. W W = 2900 J W W = 4200 J W ANSWER: Correct Part K Find the work done by the 15-newton force. Use two significant figures in your answer. Express your answer in joules. ANSWER: Correct Introduction to Potential Energy Learning Goal: Understand that conservative forces can be removed from the work integral by incorporating them into a new form of energy called potential energy that must be added to the kinetic energy to get the total mechanical energy. The first part of this problem contains short-answer questions that review the work-energy theorem. In the second part we introduce the concept of potential energy. But for now, please answer in terms of the work-energy theorem. Work-Energy Theorem The work-energy theorem states , where is the work done by all forces that act on the object, and and are the initial and final kinetic energies, respectively. Part A The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the particle if the force has a component parallel to the motion. W = -1900 J W W = -1800 J Kf = Ki + Wall Wall Ki Kf Choose the best answer to fill in the blanks above: ANSWER: Correct It is important that the force have a component acting in the direction of motion. For example, if a ball is attached to a string and whirled in uniform circular motion, the string does apply a force to the ball, but since the string's force is always perpendicular to the motion it does no work and cannot change the kinetic energy of the ball. Part B To calculate the change in energy, you must know the force as a function of _______. The work done by the force causes the energy change. Choose the best answer to fill in the blank above: ANSWER: Correct Part C To illustrate the work-energy concept, consider the case of a stone falling from to under the influence of gravity. Using the work-energy concept, we say that work is done by the gravitational _____, resulting in an increase of the ______ energy of the stone. Choose the best answer to fill in the blanks above: distance / potential distance / kinetic vertical displacement / potential none of the above acceleration work distance potential energy xi xf ANSWER: Correct Potential Energy You should read about potential energy in your text before answering the following questions. Potential energy is a concept that builds on the work-energy theorem, enlarging the concept of energy in the most physically useful way. The key aspect that allows for potential energy is the existence of conservative forces, forces for which the work done on an object does not depend on the path of the object, only the initial and final positions of the object. The gravitational force is conservative; the frictional force is not. The change in potential energy is the negative of the work done by conservative forces. Hence considering the initial and final potential energies is equivalent to calculating the work done by the conservative forces. When potential energy is used, it replaces the work done by the associated conservative force. Then only the work due to nonconservative forces needs to be calculated. In summary, when using the concept of potential energy, only nonconservative forces contribute to the work, which now changes the total energy: , where and are the final and initial potential energies, and is the work due only to nonconservative forces. Now, we will revisit the falling stone example using the concept of potential energy. Part D Rather than ascribing the increased kinetic energy of the stone to the work of gravity, we now (when using potential energy rather than work-energy) say that the increased kinetic energy comes from the ______ of the _______ energy. Choose the best answer to fill in the blanks above: ANSWER: force / kinetic potential energy / potential force / potential potential energy / kinetic Kf + Uf = Ef = Wnc + Ei = Wnc + Ki + Ui Uf Ui Wnc Correct Part E This process happens in such a way that total mechanical energy, equal to the ______ of the kinetic and potential energies, is _______. Choose the best answer to fill in the blanks above: ANSWER: Correct Problem 11.7 Part A How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: work / potential force / kinetic change / potential sum / conserved sum / zero sum / not conserved difference / conserved F  = (− + i ^ ) N j ^ ! = r m i ^ Correct Part B How much work is done by the force 2.2 6.6 on a particle that moves through displacement 3.9 Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.10 A 1.8 book is lying on a 0.80- -high table. You pick it up and place it on a bookshelf 2.27 above the floor. Part A How much work does gravity do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B W = -8.6 J F  = (− + i ^ ) N j ^ ! = r m? j ^ W = 26 J kg m m Wg = -26 J How much work does your hand do on the book? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.12 The three ropes shown in the bird's-eye view of the figure are used to drag a crate 3.3 across the floor. Part A How much work is done by each of the three forces? Express your answers using two significant figures. Enter your answers numerically separated by commas. ANSWER: WH = 26 J m W1 , W2 , W3 = 1.9,1.2,-2.1 kJ Correct Enhanced EOC: Problem 11.16 A 1.2 particle moving along the x-axis experiences the force shown in the figure. The particle's velocity is 4.6 at . You may want to review ( pages 286 - 287) . For help with math skills, you may want to review: The Definite Integral Part A What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: kg m/s x = 0m x = 2m x = 0 m x = 0 m x = 2 m x = 2 m x = 2 m Correct Part B What is its velocity at ? Express your answer to two significant figures and include the appropriate units. Hint 1. How to approach the problem What is the work–kinetic energy theorem? What is the kinetic energy at ? How is the work done in going from to related to force shown in the graph? Can the work be negative? Using the work–kinetic energy theorem, what is the kinetic energy at ? What is the velocity at ? ANSWER: Correct Work on a Sliding Block A block of weight sits on a frictionless inclined plane, which makes an angle with respect to the horizontal, as shown. A force of magnitude , applied parallel to the incline, pulls the block up the plane at constant speed. v = 6.2 ms x = 4m x = 0 m x = 0 m x = 4 m x = 4 m x = 4 m v = 4.6 ms w  F Part A The block moves a distance up the incline. The block does not stop after moving this distance but continues to move with constant speed. What is the total work done on the block by all forces? (Include only the work done after the block has started moving, not the work needed to start the block moving from rest.) Express your answer in terms of given quantities. Hint 1. What physical principle to use To find the total work done on the block, use the work-energy theorem: . Hint 2. Find the change in kinetic energy What is the change in the kinetic energy of the block, from the moment it starts moving until it has been pulled a distance ? Remember that the block is pulled at constant speed. Hint 1. Consider kinetic energy If the block's speed does not change, its kinetic energy cannot change. ANSWER: ANSWER: L Wtot Wtot = Kf − Ki L Kf − Ki = 0 Wtot = 0 Correct Part B What is , the work done on the block by the force of gravity as the block moves a distance up the incline? Express the work done by gravity in terms of the weight and any other quantities given in the problem introduction. Hint 1. Force diagram Hint 2. Force of gravity component What is the component of the force of gravity in the direction of the block's displacement (along the inclined plane)? Express your answer in terms of and . Hint 1. Relative direction of the force and the motion Remember that the force of gravity acts down the plane, whereas the block's displacement is directed up the plane. ANSWER: Wg L w w  ANSWER: Correct Part C What is , the work done on the block by the applied force as the block moves a distance up the incline? Express your answer in terms of and other given quantities. Hint 1. How to find the work done by a constant force Remember that the work done on an object by a particular force is the integral of the dot product of the force and the instantaneous displacement of the object, over the path followed by the object. In this case, since the force is constant and the path is a straight segment of length up the inclined plane, the dot product becomes simple multiplication. ANSWER: Correct Part D What is , the work done on the block by the normal force as the block moves a distance up the inclined plane? Express your answer in terms of given quantities. Hint 1. First step in computing the work Fg|| = −wsin() Wg = −wLsin() WF F L F L WF = FL Wnormal L The work done by the normal force is equal to the dot product of the force vector and the block's displacement vector. The normal force and the block's displacement vector are perpendicular. Therefore, what is their dot product? ANSWER: ANSWER: Correct Problem 11.20 A particle moving along the -axis has the potential energy , where is in . Part A What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. N  L = 0 Wnormal = 0 y U = 3.2y3 J y m y y = 0 m Fy = 0 N y y = 1 m ANSWER: Correct Part C What is the -component of the force on the particle at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.28 A cable with 25.0 of tension pulls straight up on a 1.08 block that is initially at rest. Part A What is the block's speed after being lifted 2.40 ? Solve this problem using work and energy. Express your answer with the appropriate units. ANSWER: Correct Fy = -9.6 N y y = 2 m Fy = -38 N N kg m vf = 8.00 ms Problem 11.29 Part A How much work does an elevator motor do to lift a 1500 elevator a height of 110 ? Express your answer with the appropriate units. ANSWER: Correct Part B How much power must the motor supply to do this in 50 at constant speed? Express your answer with the appropriate units. ANSWER: Correct Problem 11.32 How many energy is consumed by a 1.20 hair dryer used for 10.0 and a 11.0 night light left on for 16.0 ? Part A Hair dryer: Express your answer with the appropriate units. kg m Wext = 1.62×106 J s = 3.23×104 P W kW min W hr ANSWER: Correct Part B Night light: Express your answer with the appropriate units. ANSWER: Correct Problem 11.42 A 2500 elevator accelerates upward at 1.20 for 10.0 , starting from rest. Part A How much work does gravity do on the elevator? Express your answer with the appropriate units. ANSWER: Correct W = 7.20×105 J = 6.34×105 W J kg m/s2 m −2.45×105 J Part B How much work does the tension in the elevator cable do on the elevator? Express your answer with the appropriate units. ANSWER: Correct Part C Use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10.0 . Express your answer with the appropriate units. ANSWER: Correct Part D What is the speed of the elevator as it reaches 10.0 ? Express your answer with the appropriate units. ANSWER: Correct 2.75×105 J m 3.00×104 J m 4.90 ms Problem 11.47 A horizontal spring with spring constant 130 is compressed 17 and used to launch a 2.4 box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15. Part A Use work and energy to find how far the box slides across the rough surface before stopping. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 11.49 Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 and the coefficient of rolling friction is 0.45. Part A Use work and energy to find the length of a ramp that will stop a 15,000 truck that enters the ramp at 30 . Express your answer to two significant figures and include the appropriate units. ANSWER: Correct N/m cm kg l = 53 cm kg m/s l = 83 m Problem 11.51 Use work and energy to find an expression for the speed of the block in the following figure just before it hits the floor. Part A Find an expression for the speed of the block if the coefficient of kinetic friction for the block on the table is . Express your answer in terms of the variables , , , , and free fall acceleration . ANSWER: Part B Find an expression for the speed of the block if the table is frictionless. Express your answer in terms of the variables , , , and free fall acceleration . ANSWER: μk M m h μk g v = M m h g Problem 11.57 The spring shown in the figure is compressed 60 and used to launch a 100 physics student. The track is frictionless until it starts up the incline. The student's coefficient of kinetic friction on the incline is 0.12 . Part A What is the student's speed just after losing contact with the spring? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How far up the incline does the student go? Express your answer to two significant figures and include the appropriate units. ANSWER: v = cm kg 30 v = 17 ms Correct Score Summary: Your score on this assignment is 93.6%. You received 112.37 out of a possible total of 120 points. !s = 41 m

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